For every $a+b\sqrt{14}∈\mathbb{Z}[\sqrt{14}]$ define the norm function by: $N(a+b\sqrt{14})=|a^2-14b^2|$
Show that $\mathbb{Z}[\sqrt{14}]$ is not an Euclidean domain with regard to this norm.
My Attempt: By calculating $N(15+4\sqrt{14})=1$, we can see that $15+4\sqrt{14}\ $ is a unit in $\mathbb{Z}[\sqrt{14}]$. Also it can be shown that $4+\sqrt{14}$ is a universal side divisor. (Because it divides $2$.) Yet I don't know how these facts can help for the proof we want.
By searching the web and finding this link: Have I found an example of norm-Euclidean failure in $\mathbb Z [\sqrt{14}]$?, I saw that the pair $2, 1+\sqrt{14}$ can be used as a contradiction for $\mathbb{Z}[\sqrt{14}]$ being an Euclidean domain regarding this norm. But the proof uses the multiplicativity of norm in $\mathbb{Q}[\sqrt{14}]$ , which is not what the question necessarily wants.
So I wrote it down myself and tried to prove it this way: Suppose there exists $a,b,c,d∈\mathbb{Z}$ such that $2=(a+b\sqrt{14})(1+\sqrt{14})+c+d\sqrt{14}$ and $N(c+d\sqrt{14})<13=N(1+\sqrt{14})$
$2=(a+14b+c)+(a+b+d)\sqrt{14}⇒2=a+14b+c $ and $a+b+d=0$
But also we have:
$|c^2-14d^2|<13⇒|c^2-14d^2|=|c^2-14d^2|$(mod $13$)$=|c^2-d^2|=|(c-d)(c+d)|$
I don't know how to continue from hereon. I would really appreciate it if either someone could complete my proof or could prove it in some way or other that doesn't use the norm in $\mathbb{Q}[\sqrt{14}]$.
Note: I know that $\mathbb{Z}[\sqrt{14}]$ is actually an Euclidean domain by using another function as norm.
Thanks for your help!